Homework Help: chem--buffers

Posted by Rav on Sunday, March 7, 2010 at 5:02pm.

Calculate [H30+] in a solution that is 0.0525M HCl and 0.768M NaC2H3O2. Ka(HC2H3O2)= 1.8 x10^-5.

the way i did it is that i used the given molar of HCl to find the moles of Hydronium in the acid, and then i used that to find the moles of OH- in it, using kw. then thru the equation:

C2H3O2^- + H2O <==> HC2H3O2 + OH^-

and thru ice tables used the mols of OH- as initial and solved for x, then got OH- final, and finally found H3O+......this seems to be a way longer way the way i did it...i am sure ther's an easier way to do this...and this might also be wrong. how to do this?

chem--buffers - DrBob222, Sunday, March 7, 2010 at 5:57pm
I would be interested in the calculations you didn't show. If I understand what you did you have (H^+) 0.0525 and that makes the "initial" OH = 1.9 x 10^-13 which seems unreasonably high.
I would approach it a little differently. I would start with HCl is a strong acid and the H^+ will be tied up by the acetate (let's call that Ac^- to make typing simpler).
So if that happens, then
Ac^- + H^+ ==> HAc
I would do the ICE chart with this.
We start with (say 1 L) of 0.768 M Ac^- and 0.0525 H^+. That should combine to form 0.0525 HAc and leave essentially zero free H^+ and Ac^- of 0.768-0.0525=0.7155.
Then solve the Ka expression for H^+ OR use the Henderson-Hasselbalch equation for pH. I get something like 5.88 for pH but check me out on that.
chem--buffers - Rav, Sunday, March 7, 2010 at 9:16pm
yes, that is exactly what i get(i got 5.87, but that's probably just a simple rounding error). i used the wrong equation, and didn't include the hydrogen ion from HCl in the equation, and just used the acetic ion as the only source of the hydroxide ion. But your process does make more sense, because, as now i understand and which is correct, the hydrogen from HCl should have reacted with acetic ion.

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